Delete rows with grep() and lapply with data.table

Question

I'm learning how data tables work and I'm trying to use grep() on two columns (id1 and id2) to delete rows that don't return TRUE.

I know I have to use the function lapply() but it always returns the followed error :

argument 'pattern' has length > 1 and only the first element will be used

I tried this (and I know it's wrong) :

DT[, lapply(.SD, grepl(id1, id2)), by= id]

The data I'm working on :

structure(list(id = c(52L, 52L, 52L, 52L, 54L, 54L, 84L, 84L, 
87L, 87L, 129L, 129L, 130L, 130L, 130L), id1 = c("8113H187", 
"3505H6", "3505H6", "3505H6", "3505H6", "3505H6", "3505H6", "3505H6", 
"8113H187", "8113H187", "3505H6", "3505H6", "3505H6", "3505H6", 
"3505H6"), id2 = c("3505H6856", "3505H6856", "3505H6856", "3505H6856", 
"3505H67158", "3505H67158", "3505H63188", "3505H63188", "3505H64691", 
"3505H64691", "3505H664133", "3505H664133", "3505H658134", "3505H658134", 
"3505H658134")), .Names = c("id", "id1", "id2"), row.names = c(NA, 
-15L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x00000000064f0788>)

`grepl` pattern argument isn't vectorized. Try something like `library(stringi) ; DT[, stri_detect_fixed(id2, id1)]` — David Arenburg, Mar 20 '17 at 15:17

score 1 · Answer 1 · answered Mar 20 '17 at 15:17

1

We can use Map to do compare the corresponding elements in 'id1' as pattern to the elements in 'ID2'

DT[unlist(Map(grepl, id1, id2))]

answered Mar 20 '17 at 15:17

akrun

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score 1 · Accepted Answer · answered Mar 20 '17 at 15:26

DT[mapply( grepl, id1, id2), ]

#     id    id1         id2
# 1:  52 3505H6   3505H6856
# 2:  52 3505H6   3505H6856
# 3:  52 3505H6   3505H6856
# 4:  54 3505H6  3505H67158
# 5:  54 3505H6  3505H67158
# 6:  84 3505H6  3505H63188
# 7:  84 3505H6  3505H63188
# 8: 129 3505H6 3505H664133
# 9: 129 3505H6 3505H664133
# 10: 130 3505H6 3505H658134
# 11: 130 3505H6 3505H658134
# 12: 130 3505H6 3505H658134

Delete rows with grep() and lapply with data.table

2 Answers2